Answer
$\det (A)=14$
$\det (A^{-1})=\frac{1}{14}$
$\det (-3A)=-378$
Work Step by Step
The determinant for the given system is:
$\det (A)=1.(3-4)+1.(9-0)+2.3=14$
We have $AA^{-1}=I$
$\det(AA^{-1}=\det (I)=1$
$\det (A) . \det (A^{-1})=1$
$\det (A^{-1})=\frac{1}{\det (A)}=\frac{1}{14}$
If A is $n \times n$ matrix then,
$\det (kA)=(k)^n\det (A)$
Hence, $\det (-3A)=(-3)^3.(14)=-378$
